Question 59
In the triangle *ABC* the point *N* is the midpoint of the median *CM*. The point *X* satisfies the following conditions: $\angle XMC = \angle MBC$, $\angle XCM = \angle MCB$ and points *X* and *B* lie from different sides with respect to the line *CM*. Let $\omega$ be the circumscribed circle of $\triangle AMX$. Prove that:
Part (a). *CM* is tangent to $\omega$;
Part (b). Lines *NX* and *AC* intersect on the circle $\omega$.
Part (a). *CM* is tangent to $\omega$;
Part (b). Lines *NX* and *AC* intersect on the circle $\omega$.